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\begin{document}
\subsection*{Exercises for Section 8}
    \begin{enumerate}[start = 7,itemsep=2\baselineskip]
        \item {\begin{tabular}[t]{@{}p{.5\linewidth}l}
        $y = f(x)=\frac{1}{2}x^3$ & $x=2$ and $\Delta{x}= dx = 0.1$
        \end{tabular}
        $y' = f'(x)=\frac{3}{2}x^2$\\
        $dy = f'(x)dx = \frac{3}{2} \cdot 2^2 \cdot 0.1 = 0.6$\\
        $\Delta{y} = f(x+\Delta{x}) - f(x)= \left(\frac{1}{2}(2+0.1)^3 - \frac{1}{2}(2)^3 \right) = 0.6305$}

        \item {\begin{tabular}[t]{@{}p{.5\linewidth}l}
        $y = f(x)=1-2x^2$ & $x=0$ and $\Delta{x}= dx = -0.1$
        \end{tabular}\\
        $y' = f'(x)=-4x $\\
        $dy = f'(x)dx = -4 \cdot 0 \cdot -0.1 = 0$\\
        $\Delta{y} = f(x+\Delta{x}) - f(x)= \left((1-2(0+-0.1)^2) - (1-2(0)^2)\right) = -0.02$}

        \item {\begin{tabular}[t]{@{}p{.5\linewidth}l}
        $y = f(x)=x^4-1$ & $x=-1$ and $\Delta{x} = dx = 0.01$
        \end{tabular}\\
        $y' = f'(x)=4x^3 $\\
        $dy = f'(x)dx = 4(-1)^3 \cdot 0.01 = -0.04$\\
        $\Delta{y} = f(x+\Delta{x}) - f(x)= \left(((-1+0.01)^4 - 1) - ((-1)^4 - 1)\right) = -0.03940399$}

        \item {\begin{tabular}[t]{@{}p{.5\linewidth}l}
        $y = f(x)=2x+1$ & $x=2$ and $\Delta{x} = dx = 0.01$
        \end{tabular}\\
        $y' = f'(x)= 2 $\\
        $dy = f'(x)dx = 2 \cdot 0.01 = 0.02$\\
        $\Delta{y} = f(x+\Delta{x}) - f(x)= \left( 2(2+0.01)+1 - 2\cdot2+1 \right) = 0.02$}
    \end{enumerate}

\end{document}
